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2

$begingroup$

$a neq a$

$a$ is the set whose only element is the a (and no others). $a$ is the set whose only element is the set $a$.

Does this mean the 'element a' is not equal to 'set $a$'?

algebra-precalculus

share|cite|improve this question

edited Mar 17 at 14:26

Max

9191319

asked Mar 17 at 14:16

Chen YunChen Yun

133

$endgroup$

• 
  
  
  

  
  7
  

  
  
  
  
  
  

  
  $begingroup$
  Indeed, those are not equal.
  $endgroup$
  – StackTD
  Mar 17 at 14:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Related: Why is $1$ not equal to $1,1$?, and Is $emptyset$ a subset of $emptyset$
  $endgroup$
  – JMoravitz
  Mar 17 at 14:21
  
  

  
  
  
  

add a comment | 

2

$begingroup$

$a neq a$

$a$ is the set whose only element is the a (and no others). $a$ is the set whose only element is the set $a$.

Does this mean the 'element a' is not equal to 'set $a$'?

algebra-precalculus

share|cite|improve this question

edited Mar 17 at 14:26

Max

9191319

asked Mar 17 at 14:16

Chen YunChen Yun

133

$endgroup$

• 
  
  
  

  
  7
  

  
  
  
  
  
  

  
  $begingroup$
  Indeed, those are not equal.
  $endgroup$
  – StackTD
  Mar 17 at 14:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Related: Why is $1$ not equal to $1,1$?, and Is $emptyset$ a subset of $emptyset$
  $endgroup$
  – JMoravitz
  Mar 17 at 14:21
  
  

  
  
  
  

add a comment | 

$begingroup$

$a neq a$

$a$ is the set whose only element is the a (and no others). $a$ is the set whose only element is the set $a$.

Does this mean the 'element a' is not equal to 'set $a$'?

algebra-precalculus

share|cite|improve this question

edited Mar 17 at 14:26

Max

9191319

asked Mar 17 at 14:16

Chen YunChen Yun

133

$endgroup$

share|cite|improve this question

edited Mar 17 at 14:26

Max

9191319

asked Mar 17 at 14:16

Chen YunChen Yun

133

share|cite|improve this question

edited Mar 17 at 14:26

Max

9191319

asked Mar 17 at 14:16

Chen YunChen Yun

133

edited Mar 17 at 14:26

Max

9191319

edited Mar 17 at 14:26

Max

9191319

asked Mar 17 at 14:16

Chen YunChen Yun

133

asked Mar 17 at 14:16

Chen YunChen Yun

133

3 Answers
3

active

oldest

votes

1

$begingroup$

Even though people sometimes get sloppy about it, $a$ and $a$ are not the same object. $a$ is the only element of the set $a$.

share|cite|improve this answer

answered Mar 17 at 14:20

KlausKlaus

2,792113

$endgroup$

add a comment | 

0

$begingroup$

They are not equal.

Intuitively, $a$ means a set which contains an element $a$; while $a$ means a set that contains a set $a$ as its element.

From ZFC axiom: Every non-empty set $x$ contains a member $y$ such that $x$ and $y$ are disjoint sets.

share|cite|improve this answer

answered Mar 17 at 14:33

Yujie ZhaYujie Zha

6,98611729

$endgroup$

add a comment | 

0

$begingroup$

In general: $$x=yiff x=y$$

Then we can conclude that also:$$xneqyiff xneq y$$

Applying that in your case we find that the statement $aneqa$ is the same statement as $aneqa$.

---

Sidenote:

If also the axiom of regularity is accepted then this statement is true for every $a$.

This because on base of that axiom it can be proved that $anotin a$ is true for every $a$ while $a=a$ implies that $ain a$.

share|cite|improve this answer

edited Mar 17 at 16:13

answered Mar 17 at 14:45

drhabdrhab

103k545136

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  From Wikipedia, axiom of extensionality: "what the axiom is really saying is that two sets are equal if and only if they have precisely the same members. The essence of this is: A set is determined uniquely by its members." >> a and a are not same.
  $endgroup$
  – Chen Yun
  Mar 17 at 15:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Axiom of regularity: "No set is an element of itself" "We see that there must be an element of A which is disjoint from A. Since the only element of A is A, it must be that A is disjoint from A. So, since A ∈ A, we cannot have A ∈ A (by the definition of disjoint)." >> I don't get your meaning, I assume as a is an element of a, they are not going to be equal. Is this correct??
  $endgroup$
  – Chen Yun
  Mar 17 at 15:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What you mention ("No set is an element of itself") is not the axiom of regularity itself but is a consequence of the axiom of regularity. The axiom says that $A$ must have an element $x$ with $xcapA=varnothing$. The only candidate for $x$ is $A$ so that we can conclude that $AcapA=varnothing$. From this it follows that $Anotin A$. If that's what you are saying then I fully agree with you and based on the axiom it has been proved that $ain a$ cannot be a true statement (as stated in my answer). "I don't get your meaning..." What meaning?
  $endgroup$
  – drhab
  Mar 17 at 16:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What I said about the axiom of extensionality in my answer was wrong and I removed it.
  $endgroup$
  – drhab
  Mar 17 at 16:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh I can't get the false statement "a=a implies that a∈a". From your comment above "A∩A=∅. From this it follows that A∉A". >> No matter what is the element surely it is not equal to the set because they have no intersection of each other, disjoint. Thus we could not find "element of..." within it.
  $endgroup$
  – Chen Yun
  Mar 17 at 17:15
  

  
  
  
  

 | 
show 1 more comment

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1

$begingroup$

Even though people sometimes get sloppy about it, $a$ and $a$ are not the same object. $a$ is the only element of the set $a$.

share|cite|improve this answer

answered Mar 17 at 14:20

KlausKlaus

2,792113

$endgroup$

add a comment | 

1

$begingroup$

Even though people sometimes get sloppy about it, $a$ and $a$ are not the same object. $a$ is the only element of the set $a$.

share|cite|improve this answer

answered Mar 17 at 14:20

KlausKlaus

2,792113

$endgroup$

add a comment | 

$begingroup$

Even though people sometimes get sloppy about it, $a$ and $a$ are not the same object. $a$ is the only element of the set $a$.

share|cite|improve this answer

answered Mar 17 at 14:20

KlausKlaus

2,792113

$endgroup$

share|cite|improve this answer

answered Mar 17 at 14:20

KlausKlaus

2,792113

answered Mar 17 at 14:20

KlausKlaus

2,792113

answered Mar 17 at 14:20

KlausKlaus

2,792113

0

$begingroup$

They are not equal.

Intuitively, $a$ means a set which contains an element $a$; while $a$ means a set that contains a set $a$ as its element.

From ZFC axiom: Every non-empty set $x$ contains a member $y$ such that $x$ and $y$ are disjoint sets.

share|cite|improve this answer

answered Mar 17 at 14:33

Yujie ZhaYujie Zha

6,98611729

$endgroup$

add a comment | 

0

$begingroup$

They are not equal.

Intuitively, $a$ means a set which contains an element $a$; while $a$ means a set that contains a set $a$ as its element.

From ZFC axiom: Every non-empty set $x$ contains a member $y$ such that $x$ and $y$ are disjoint sets.

share|cite|improve this answer

answered Mar 17 at 14:33

Yujie ZhaYujie Zha

6,98611729

$endgroup$

add a comment | 

$begingroup$

They are not equal.

Intuitively, $a$ means a set which contains an element $a$; while $a$ means a set that contains a set $a$ as its element.

From ZFC axiom: Every non-empty set $x$ contains a member $y$ such that $x$ and $y$ are disjoint sets.

share|cite|improve this answer

answered Mar 17 at 14:33

Yujie ZhaYujie Zha

6,98611729

$endgroup$

share|cite|improve this answer

answered Mar 17 at 14:33

Yujie ZhaYujie Zha

6,98611729

answered Mar 17 at 14:33

Yujie ZhaYujie Zha

6,98611729

answered Mar 17 at 14:33

Yujie ZhaYujie Zha

6,98611729

0

$begingroup$

In general: $$x=yiff x=y$$

Then we can conclude that also:$$xneqyiff xneq y$$

Applying that in your case we find that the statement $aneqa$ is the same statement as $aneqa$.

---

Sidenote:

If also the axiom of regularity is accepted then this statement is true for every $a$.

This because on base of that axiom it can be proved that $anotin a$ is true for every $a$ while $a=a$ implies that $ain a$.

share|cite|improve this answer

edited Mar 17 at 16:13

answered Mar 17 at 14:45

drhabdrhab

103k545136

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  From Wikipedia, axiom of extensionality: "what the axiom is really saying is that two sets are equal if and only if they have precisely the same members. The essence of this is: A set is determined uniquely by its members." >> a and a are not same.
  $endgroup$
  – Chen Yun
  Mar 17 at 15:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Axiom of regularity: "No set is an element of itself" "We see that there must be an element of A which is disjoint from A. Since the only element of A is A, it must be that A is disjoint from A. So, since A ∈ A, we cannot have A ∈ A (by the definition of disjoint)." >> I don't get your meaning, I assume as a is an element of a, they are not going to be equal. Is this correct??
  $endgroup$
  – Chen Yun
  Mar 17 at 15:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What you mention ("No set is an element of itself") is not the axiom of regularity itself but is a consequence of the axiom of regularity. The axiom says that $A$ must have an element $x$ with $xcapA=varnothing$. The only candidate for $x$ is $A$ so that we can conclude that $AcapA=varnothing$. From this it follows that $Anotin A$. If that's what you are saying then I fully agree with you and based on the axiom it has been proved that $ain a$ cannot be a true statement (as stated in my answer). "I don't get your meaning..." What meaning?
  $endgroup$
  – drhab
  Mar 17 at 16:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What I said about the axiom of extensionality in my answer was wrong and I removed it.
  $endgroup$
  – drhab
  Mar 17 at 16:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh I can't get the false statement "a=a implies that a∈a". From your comment above "A∩A=∅. From this it follows that A∉A". >> No matter what is the element surely it is not equal to the set because they have no intersection of each other, disjoint. Thus we could not find "element of..." within it.
  $endgroup$
  – Chen Yun
  Mar 17 at 17:15
  

  
  
  
  

 | 
show 1 more comment

0

$begingroup$

In general: $$x=yiff x=y$$

Then we can conclude that also:$$xneqyiff xneq y$$

Applying that in your case we find that the statement $aneqa$ is the same statement as $aneqa$.

---

Sidenote:

If also the axiom of regularity is accepted then this statement is true for every $a$.

This because on base of that axiom it can be proved that $anotin a$ is true for every $a$ while $a=a$ implies that $ain a$.

share|cite|improve this answer

edited Mar 17 at 16:13

answered Mar 17 at 14:45

drhabdrhab

103k545136

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  From Wikipedia, axiom of extensionality: "what the axiom is really saying is that two sets are equal if and only if they have precisely the same members. The essence of this is: A set is determined uniquely by its members." >> a and a are not same.
  $endgroup$
  – Chen Yun
  Mar 17 at 15:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Axiom of regularity: "No set is an element of itself" "We see that there must be an element of A which is disjoint from A. Since the only element of A is A, it must be that A is disjoint from A. So, since A ∈ A, we cannot have A ∈ A (by the definition of disjoint)." >> I don't get your meaning, I assume as a is an element of a, they are not going to be equal. Is this correct??
  $endgroup$
  – Chen Yun
  Mar 17 at 15:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What you mention ("No set is an element of itself") is not the axiom of regularity itself but is a consequence of the axiom of regularity. The axiom says that $A$ must have an element $x$ with $xcapA=varnothing$. The only candidate for $x$ is $A$ so that we can conclude that $AcapA=varnothing$. From this it follows that $Anotin A$. If that's what you are saying then I fully agree with you and based on the axiom it has been proved that $ain a$ cannot be a true statement (as stated in my answer). "I don't get your meaning..." What meaning?
  $endgroup$
  – drhab
  Mar 17 at 16:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What I said about the axiom of extensionality in my answer was wrong and I removed it.
  $endgroup$
  – drhab
  Mar 17 at 16:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh I can't get the false statement "a=a implies that a∈a". From your comment above "A∩A=∅. From this it follows that A∉A". >> No matter what is the element surely it is not equal to the set because they have no intersection of each other, disjoint. Thus we could not find "element of..." within it.
  $endgroup$
  – Chen Yun
  Mar 17 at 17:15
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

In general: $$x=yiff x=y$$

Then we can conclude that also:$$xneqyiff xneq y$$

Applying that in your case we find that the statement $aneqa$ is the same statement as $aneqa$.

---

Sidenote:

If also the axiom of regularity is accepted then this statement is true for every $a$.

This because on base of that axiom it can be proved that $anotin a$ is true for every $a$ while $a=a$ implies that $ain a$.

share|cite|improve this answer

edited Mar 17 at 16:13

answered Mar 17 at 14:45

drhabdrhab

103k545136

$endgroup$

share|cite|improve this answer

edited Mar 17 at 16:13

answered Mar 17 at 14:45

drhabdrhab

103k545136

answered Mar 17 at 14:45

drhabdrhab

103k545136

answered Mar 17 at 14:45

drhabdrhab

103k545136